Two positively charged particles are separated on an axis by 0.1 meters; q1 has a 3 microcoulomb charge and q2 has a 2 microcoul
omb charge. Recall that k = 8.99 × 10^9 N • meters squared over Coulombs squared.
a) What is the magnitude of the force applied between q1 and q2?
b) Is the force on q2 directed toward or away from q1?
please help!
a) What is the magnitude of the force applied between q1 and q2?
b) Is the force on q2 directed toward or away from q1?
please help!
1 answer:
You might be interested in
Answer:
average value of the resulting force
Explanation:
The average module value of this resulting force is equivalent to 2.0. 10⁵ N.
The impulse of a force can be calculated by the product of the intensity of the force applied by the time interval in which it is applied -
I = F.Δt
Where,
F = Strength in Newtons
Δt = time interval in seconds
I = Impulse in N.s
The impulse of a force is equivalent to the variation of the amount of movement it causes in the body.
I = ΔQ
The amount of movement is a vector quantity that results from the multiplication of the mass of a body by its speed. Its direction and direction are the same as the velocity vector of the body.
Q = m-V
As the car goes to rest after the application of force, the amount of final movement of the car is equivalent to zero.
I = 0 - mV
F. Δt = - mV
F. 0,1 = - 1000. 20
F = - 20000/0,1
F = 200,000 N
F = 2,0. 10⁵ N
